Which between $\ce{HI}$ and $\ce{HF}$ has a greater dipolar moment? I think it is $\ce{HI}$ because the atomic radius of $\ce{I}$ is greater.

  • 1
    $\begingroup$ How much charge separation is between the atoms on each compound? What is the difference between electronegativities of H and I? H and F? $\endgroup$ Commented Jan 4, 2018 at 13:29
  • $\begingroup$ F is more electronegative than I, how do I evalue if it compensates for the smaller distance? $\endgroup$
    – Mattiatore
    Commented Jan 4, 2018 at 13:36
  • $\begingroup$ One method is to look up the "rulr" for ionic character versus electronegativity difference, see the plot in chemistry.stackexchange.com/questions/9222/…. Use the curve to figure out where HI might be. Which is the bigger factor now, atomic size or charge separation? $\endgroup$ Commented Jan 4, 2018 at 14:08
  • $\begingroup$ see chemistry.stackexchange.com/questions/31049/… $\endgroup$
    – Mithoron
    Commented Jan 4, 2018 at 16:18

1 Answer 1


The dipole moment of $\ce{HF}$ is greater than $\ce{HI}$:

  • $\ce{HF} = \pu{1.91 D}$
  • $\ce{HI} = \pu{0.42 D}$

This is due to high electronegativity of fluorine. Hence from $\mu = \vec{q} \cdot \vec{d}$, the charge of fluorine is larger than iodine, but the bond length changes only a small amount.

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    $\begingroup$ How exactly is the charge related to the electronegativity? $\endgroup$ Commented Jan 4, 2018 at 15:02
  • $\begingroup$ Higher the charge per unit area on atomic surface more is the tendency of that atom to attract electrons. (Allred Rochow Electronegativity). $\endgroup$
    – user57147
    Commented Jan 4, 2018 at 15:33

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